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olga55 [171]
4 years ago
14

List the theorems for finding zeros of higher degree polynomials

Mathematics
1 answer:
astra-53 [7]4 years ago
8 0
The Cold War 1945-<span>The Cold War 1945-1970</span>
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For any real number x,x2&gt;x3
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No. That's only true if x is less than ' 1 ', including anything negative. If x is more than +1, then the square is LESS than the cube.
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3 years ago
What is the answer to this question ?
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Answer:

4

Step-by-step explanation:

h(7)= 7²-1= 49-1= 48

f(h(7))= 48/3 -12

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How many sprinkles on a cookie
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The water company that serves St. George Island charges a base facility fee of $32.00 each month. In addition, they charge $6.53
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4 years ago
Find an equation of the line that passes through the points (-5, -3) and (3, 1)
Mkey [24]

\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-3})\quad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-5)}}}\implies \cfrac{1+3}{3+5}\implies \cfrac{4}{8}\implies \cfrac{1}{2}

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{1}{2}}[x-\stackrel{x_1}{(-5)}]\implies y+3=\cfrac{1}{2}(x+5) \\\\\\ y+3=\cfrac{1}{2}x+\cfrac{5}{2}\implies y=\cfrac{1}{2}x+\cfrac{5}{2}-3\implies y = \cfrac{1}{2}x-\cfrac{1}{2}

6 0
4 years ago
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